Question

What is `pH` of a solution prepared from `100*cm^3` of `0.51*mol*L^-1` acetic acid, and a `4.1*g` mass of sodium acetate?

Answer 1

Well see the old buffer equation..... and we get `pH~~4.76`

Explanation:

The buffer equation holds that ....`pH=pK_a+log_10{[[AcO^-]]/[[HOAc]]}`.

And so we gots to work out the concentrations of acetate ion, and acetic acid....

`[HOAc]=0.51*mol*L^-1`

`[Na^(+)""^(-)OAc]=((4.1*g)/(82.03*g*mol^-1))/(100*cm^3xx10^-3*L*cm^-3)=0.500*mol*L^-1`

...and since the `[""^(-)OAc]~~[HOAc]`, the `pH` will be near as dammit to the `pK_a` of the acid (do you follow?). Mind you, you did not quote `pK_a` for acetic acid....I happen to know that it is `4.76`...

`pH=4.76+log_10((0.50*mol*L^-1)/(0.51*mol*L^-1))=4.75`....

Note that when a buffer is prepared WITH EQUAL concentrations of the weak acid and its conjugate base, as predicted by the given equation, `pH=pK_a` necesarily. because `log_(10)1=0`